Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is the question which i am attempting to solve, and it seems to difficult to get rid of the exponents.

Show that a the two cubic curves $Y^3 = X^2 + X^3$ and $X^3 = Y^2 + Y^3$ intersect in nine points.

Any help would be appreciated. By the way i would also like to know whether is there any general method for solving such problems.

share|improve this question
3  
Groebner bases. –  Robin Chapman Oct 1 '10 at 20:16
    
@Robin: Isn't an elementary way possible. I haven't encountered Algebraic Geometry. –  anonymous Oct 1 '10 at 20:18
    
If you want a general method, then Groebner bases is it. –  Robin Chapman Oct 1 '10 at 20:19
    
@Robin Chapman: Ah, ok –  anonymous Oct 1 '10 at 20:20
    
Should the tag be ALGEBRAIC-GEOMETRY rather than ANALYTIC-GEOMETRY? –  anon Oct 2 '10 at 17:39
add comment

3 Answers

up vote 4 down vote accepted

In addition to the usual generalities on Bezout's theorem and resultants, notice that for this system, $X^2 + Y^2 = 0$, so that there are 6 finite solutions (3 on each line $X = \pm iY$, organized as 1 transverse intersection point per line, plus one double point at (0,0) on each line, leading to a quadruple intersection of the curves at 0) and accounting for the other 3 solutions, the solutions "at infinity" with $X^3=Y^3$, i.e., the asymptotic directions $(X/Y) \to \omega, \omega^3=1, |X| \to \infty$. In other words, this example is built to show the need for complex projective geometry, and the counting of points according to their multiplicities, as the environment where the Bezout 3x3=9 calculation works.

(the generalities are: if f(x,y)=0 and g(x,y)=0 are the polynomials, you can write down a basis for k[x,y]/(f=g=0), and check that as a vector space it is of dimension 9. If you meant that there are 9 distinct intersection points, you can check that by writing down the action of the multiply-by-x operator on this vector space and seeing whether it has repeated roots. This can all be done by hand but is not as efficient as what is done in computer algebra systems. )

share|improve this answer
    
I still think I'm right -- there are only three solutions to this system of equations. If you claim that there are six, or nine, then what are they explicitly? –  Laurent Lessard Oct 2 '10 at 5:11
3  
Nine. 3 points at infinity, plus the 2 points with imaginary coordinates that you calculated, plus a quadruple point at (0,0). The organization within the lines is 1 transversal point + 2 double points, not 2+1 as written in the earlier version of the posting. Taking multiplicity of the intersections into account, your "three" points represent 2 + 4 = 6 intersections. –  T.. Oct 2 '10 at 6:14
    
Also, locally near 0 the equations look like X^2=Y^2=0. This is the intersection of two double lines, which are the tangent lines at the respective cusps of the curves (plot the graph to see that there are cusps at (0,0) ). The 4-fold intersection is a degeneration of the intersection of one pair of lines with another pair, so the multiplicity should be thought of as 2x2. –  T.. Oct 2 '10 at 6:35
add comment

Like your earlier question about how to prove the associativity of the group law on elliptic curves, this too is a consequence of general results on intersection theory, esp. (variants of) Bezout's theorem. For computational purposes you can employ various constructive elimination techniques, e.g Gröbner bases, various triangularization methods, etc, which are available in most computer algebra systems

share|improve this answer
add comment

Suppose $(x,y)$ satisfies both equations. Substituting one equation into the other, we obtain $x^2+y^2=0$. The only real solution to this equation are $x=y=0$, as pointed out by Moron.

Now suppose $x\neq 0$, and so $x$ is a complex number. Then $y = \pm i x$. Eliminating $y$ from the other equation yields two possibilities:

  1. $y = ix$, and $-ix^3 = x^2 + x^3$. We assume $x\neq 0$, so we have: $-ix = 1 + x$. This yields the solution: $x = (-1+i)/2, y=(-1-i)/2$.

  2. $y = -ix$, and $ix^3 = x^2 + x^3$. Similarly to before, we have: $ix = 1+x$. This yields the solution: $x = (-1-i)/2, y=(-1+i)/2$.

As far as I can see, these are the only possible solutions... There are three solutions, not nine.

share|improve this answer
6  
There are three "set theoretic" solutions in C^2, that is, ignoring multiplicity and points at infinity. There are 9 geometric intersections in CP^2. The reason the latter form of counting is standard is that it is invariant under deformations and degenerations, as long as the curves are kept distinct so that the number of intersections is finite. –  T.. Oct 2 '10 at 6:27
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.